Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{z - 2}{z + 2} \times \dfrac{-z^2 + 6z + 16}{z^2 - 4z + 4} $
First factor out any common factors. $r = \dfrac{z - 2}{z + 2} \times \dfrac{-(z^2 - 6z - 16)}{z^2 - 4z + 4} $ Then factor the quadratic expressions. $r = \dfrac {z - 2} {z + 2} \times \dfrac {-(z + 2)(z - 8)} {(z - 2)(z - 2)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {(z - 2) \times -(z + 2)(z - 8) } {(z + 2) \times (z - 2)(z - 2) } $ $r = \dfrac {-(z + 2)(z - 8)(z - 2)} {(z - 2)(z - 2)(z + 2)} $ Notice that $(z - 2)$ and $(z + 2)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-(z + 2)(z - 8)\cancel{(z - 2)}} {\cancel{(z - 2)}(z - 2)(z + 2)} $ We are dividing by $z - 2$ , so $z - 2 \neq 0$ Therefore, $z \neq 2$ $r = \dfrac {-\cancel{(z + 2)}(z - 8)\cancel{(z - 2)}} {\cancel{(z - 2)}(z - 2)\cancel{(z + 2)}} $ We are dividing by $z + 2$ , so $z + 2 \neq 0$ Therefore, $z \neq -2$ $r = \dfrac {-(z - 8)} {z - 2} $ $ r = \dfrac{-(z - 8)}{z - 2}; z \neq 2; z \neq -2 $